Problem: Simplify and expand the following expression: $ \dfrac{1}{t + 6}- \dfrac{3}{t + 5}+ \dfrac{t}{t^2 + 11t + 30} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor the quadratic in the third term: $ \dfrac{t}{t^2 + 11t + 30} = \dfrac{t}{(t + 6)(t + 5)}$ Now we have: $ \dfrac{1}{t + 6}- \dfrac{3}{t + 5}+ \dfrac{t}{(t + 6)(t + 5)} $ The least common multiple of the denominators is: $ (t + 6)(t + 5)$ In order to get the first term over $(t + 6)(t + 5)$ , multiply by $\dfrac{t + 5}{t + 5}$ $ \dfrac{1}{t + 6} \times \dfrac{t + 5}{t + 5} = \dfrac{t + 5}{(t + 6)(t + 5)} $ In order to get the second term over $(t + 6)(t + 5)$ , multiply by $\dfrac{t + 6}{t + 6}$ $ \dfrac{3}{t + 5} \times \dfrac{t + 6}{t + 6} = \dfrac{3(t + 6)}{(t + 6)(t + 5)} $ Now we have: $ \dfrac{t + 5}{(t + 6)(t + 5)} - \dfrac{3(t + 6)}{(t + 6)(t + 5)} + \dfrac{t}{(t + 6)(t + 5)} $ $ = \dfrac{ t + 5 - 3(t + 6) + t} {(t + 6)(t + 5)} $ Expand: $ = \dfrac{t + 5 - 3t - 18 + t}{t^2 + 11t + 30} $ $ = \dfrac{-t - 13}{t^2 + 11t + 30}$